JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 17)
A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :
6 V/m
10 V/m
30 V/m
24 V/m
Explanation
Prefracted = $${{96} \over {100}}Pi$$
$$ \Rightarrow $$ K2A$$Pi_t^2$$ = $${{96} \over {100}}$$ K1A$$_i^2$$
$$ \Rightarrow $$ r2A$$_i^2$$ = $${{96} \over {100}}$$ r1A$$_i^2$$
$$ \Rightarrow $$ A$$_t^2$$ = $${{96} \over {100}} \times {1 \over {{3 \over 2}}} \times {\left( {30} \right)^2}$$
A1$$\sqrt {{{64} \over {100}} \times {{\left( {30} \right)}^2}} = 24$$
$$ \Rightarrow $$ K2A$$Pi_t^2$$ = $${{96} \over {100}}$$ K1A$$_i^2$$
$$ \Rightarrow $$ r2A$$_i^2$$ = $${{96} \over {100}}$$ r1A$$_i^2$$
$$ \Rightarrow $$ A$$_t^2$$ = $${{96} \over {100}} \times {1 \over {{3 \over 2}}} \times {\left( {30} \right)^2}$$
A1$$\sqrt {{{64} \over {100}} \times {{\left( {30} \right)}^2}} = 24$$
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